Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Using Cartesian Coordinates we mark a point on a graph by how far along and how far up it is.
The left $-$ right $($horizontal$)$ direction is commonly called $X-$ axis.
The up $-$ down $($vertical$)$ direction is commonly called $Y-$ axis.
When we include negative values, the $x$ and $y$ axes divide the space up into $4$ pieces.
Read the information given above and below:
Two friends Veena and Arun work in the same office in Delhi. In the Christmas vacations, both decided to go
their hometowns represented by Town $A$ and Town $B$ respectively in the figure given below. Town A and Town $B$ are connected by trains from the same station $C\ ($in the given figure$)$ in Delhi.

$i$. Who will travel more distance to reach their home? $(1)$
$ii.$ Find the location of the station. $(1)$
$iii$. Find in which ratio $Y-$ axis divide Town $B$ and Station. $(2)$
OR
Find the distance between Town $A$ and Town $B. (2)$

Answer

Distance travelled by veena $=\sqrt{1-(-4)^2+(7-4)^2}$
$=\sqrt{5^2+3^2}$
$=\sqrt{25+9}$
$=\sqrt{34}$

Distance travelled by Arun $=\sqrt{(4-(-4))^2+(2-4)^2}$
$=\sqrt{64+4}$
$=\sqrt{68}$
$\therefore$ Arun will travel more distance to reach his home.
$ii$. Location of station $= (-4, 4)$

Let $y -$ axis divides station $(c)$ and Town $B$ in $K : 1$
$0=\frac{4 k-4}{k+1}$
$4 k=4$
$k=1$
$\therefore y-$ axis divides in $1: 1$
OR

$ AB =\sqrt{(4-1)^2+(2-7)^2}$
$=\sqrt{9+25}$
$=\sqrt{34}$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
The discus throw is an event in which an athlete attempts to throw a discus. The athlete spins anti$-$clockwise around one and a half times through a circle, then releases the throw. When released, the discus travels along tangent to the circular spin orbit.

In the given figure, $AB$ is one such tangent to a circle of radius $75 \ cm$ . Point $O$ is centre of the circle and $\angle ABO$
$=30^{\circ} . PQ$ is parallel to $OA.$

i. find the length of $AB. (1)$
ii. find the length of $OB. (1)$
iii. find the length of $AP. (2)$
OR
find the length of $PQ. (2)$

Answer

$\text { i. }\tan 30^{\circ} =\frac{1}{\sqrt{3}}=\frac{75}{AB}$
$\Rightarrow AB =75 \sqrt{3} \ cm$
$ii. \sin 30^{\circ}=\frac{1}{2}=\frac{75}{O B}$$\Rightarrow OB=150 \ cm$
$iii. QB =150-75=75 \ cm$
$\Rightarrow Q$ is mid point of $OB$
Since $P Q \| A O$ therefore $P$ is mid pint of $AB$
Hence AP $=\frac{75 \sqrt{3}}{2} \ cm$.
$OR$
$QB=150-75=75 \ cm$
Now, $\triangle BQP \sim \triangle BOA$
$\Rightarrow \frac{Q B}{O B}=\frac{P Q}{O A}$
$\Rightarrow \frac{1}{2}=\frac{P Q}{75}$
$\Rightarrow PQ=\frac{75}{2} \ cm$

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
Two schools $P$ and $Q$ decided to award prizes to their students for two games of Hockey $₹ x$ per student and Cricket $₹ y$ per student. School $P$ decided to award a total of $₹ 9,500$ for the two games to $5$ and $4$ students respectively; while school $Q$ decided to award $₹ 7,370$ for the two games to $4 $ and $3$ students respectively.

$i$. Represent the following information algebraically $($in terms of $x$ and $y). (1)$
$ii$. What is the prize amount for hockey? $(1)$
$iii.$ Prize amount on which game is more and by how much? $(2)$
OR
What will be the total prize amount if there are $2$ students each from two games? $(2)$

Answer

$i$. Given, prize amount for Hockey $₹ x$ and $₹ y$ for cricket per student
$\therefore$ Algebraic equations are $5 x+4 y=9500 \ldots (i)$
and $4 x+3 y=7370$
$ii$. Given, prize amount for Hockey $₹ x$ and $₹ y$ for cricket per student
$\therefore$ Algebraic equations are
$5 x+4 y=9500 \ldots (i)$
and $4 x+3 y=7370$
Multiply by $3$ in equation $(i)$ and by $4$ in equation $(ii)$
$15 x+12 y=28,500$
$16 x+12 y=29480$
On subtracting equation $(iii)$ from equation $(iv),$ we get
$x=980$
$\therefore$ Prize amount for hockey $=₹ 980$
$iii$. Given, prize amount for Hockey $₹ x$ and $₹ y$ for cricket per student
$\therefore$ Algebraic equations are
$5 x+4 y=9500 \ldots(i)$
and $ 4 x+3 y=7370$
Now, put this value in equation $(i),$ we get
$5 \times 980+4 y=9500$
$\Rightarrow 4 y=9500-4900=4600$
$\Rightarrow y=1150$
$\therefore$ Prize amount for cricket $=₹ 1150$
Difference $=1150-980=170$
$\therefore$ Prize amount for cricket is $₹ 170$ more than hockey.
OR
Total prize amount for $2$ students each from two games
$= 2 x + 2y$
$= 2 (x + y)$
$= 2 (980 + 1150)$
$= 2 \times 2130$
$= ₹ 4260$

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Maths Case study (4 Marks)

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